Question

Solve 3x – 2y + z = -5 using matrices.
-2x + 3y – 3z = 12
3x – 2y – 2z = 4

(1, 0, -3)


(1, -3, 0)


(0, 1, -3)


(0, -3, 1)

Could you help me figure out which one of these is correct please? Thanks!

Answer 1

Set up the augmented matrix!

Answer 2

3x – 2y + z = -5
-2x + 3y – 3z = 12
3x – 2y – 2z = 4

3 -2 1 -5
-2 3 -3 12
3 -2 -2 4

Answer 3

The bottom part is the augmented matrix do you see how I did it?

Answer 4

Just row reduce the augmented matrix into it's reduced echelon form.

Answer 5

sort of

Answer 6

Well I just set up the equations into it's columns and took away the variables. See how nothing has changed except that the augmented matrix has no variables.

Answer 7

oh okay

Answer 8

Now you reduce the matrix by adding rows together.

Here is an example of reduce echelon form if you don't know what it is.
http://en.wikipedia.org/wiki/Row_echelon_form

Answer 9

Can you solve it now?

Answer 10

with these resourses I bet I will be able to!

Answer 11

To tell you a little bit more information on how I got the augmented matrix
3x – 2y + z = -5
-2x + 3y – 3z = 12
3x – 2y – 2z = 4

is the same as if we rewrite it as
\[\left[\begin{matrix}3 & -2 & 1 \\ -2 & 3 & -3 \\ 3 & -2 & -2\end{matrix}\right] \left[\begin{matrix}x \\ y \\ z \end{matrix}\right] = \left[\begin{matrix}-5 \\ 12 \\ 4 \end{matrix}\right]\]

If you do matrix multiplication then you basically get back the set of equations.

You want to get reduced echelon form because once you do and you do the matrix multiplication then you will result in a set of equations where each equation has one variable or in other words you have solved for one variable.

Answer 12

If you forget you can always come back to this question but if you are still confused then ask another question.

Answer 13

dpalnc, do you have any ideas?

Answer 14

Well I pretty much know the answer....what part are you confused about?

Answer 15

oh, I thought you coudn't figure it out either. I'm just confused about the last step.

Answer 16

3 -2 1 -5
-2 3 -3 12
3 -2 -2 4

Like I said before this is your augmented matrix. You have to row reduce it to echelon form in order to solve for x y z

Echelon form basically means is to try and get each column with one number.

3 -2 1 -5 3 -2 1 -5
-2 3 -3 12 --> -2 3 -3 12 In this part I added the negative of the first
3 -2 -2 4 0 0 -3 9 row to the third one

3 -2 1 -5
-2 3 -3 12 Now I want to get rid of the -2 of the first column...
0 0 -3 9

Any ideas on how to do this?

Answer 17

do i need to find the inverse of the matrix?

Answer 18

We can use inverse matrices to solve it but if we do we have to check that the matrix is indeed invertable. Once we do we have to use the invertable matrix formula to figure out what the inverse of the matrix is. Once we find what it is we multiply the inverse to both sides.

The hard part about this is you have to find the determinant of the matrix and the adjugate which requires more work than simply solving the augmented matrix.

Answer 19

i see, what is the less time consuming way?

Answer 20

augmented matrix. But if you don't even know how to solve the augmented matrix then I doubt that you will be able to solve it using the inverse of the matrix. You learn that way more later in your book.

Answer 21

Go read your book. It's too tedious to show you how to do it over the internet.

Answer 22

Could you show your work all the way to the answer so I can look at the whole problem and see how it is done? This was the reason for my posting this question.

Answer 23

Sorry maybe later. took me 10 minutes to write that one matrix and you have to write at least 5 matrices. to do it.

Answer 24

i understand